regular expressions to replace money pattern - php

Question

I would like to ask for help with this, I wanted to check for a string
matching a decimal followed by a letter (15M, 15.3M, 15T or 15.3T) and
replace it with zeroes.
Like these:
15M -> 15000000
15.3M -> 15300000
15T -> 15000
15.3T -> 15300
I tried doing this with str_replace but can't get it right. Hope
someone can help me.

Answer 1

"T" could be "thousand" or "trillion", you know.
$s = "15.7T";
$factors = Array('T' => 1000, 'M' => 1000000, 'B' => 1000000000.);
$matches = Array();
if (0 < preg_match('/([0-9]+(?:\.[0-9]*)?)([TMB])/', $s, $matches)) {
print("$s -> " . $matches[1] * $factors[$matches[2]]);
}
prints:
15.7T -> 15700
edit:
Anchoring (makes any garbage on the front or back mean no match):
'/^(...)$/'
You may want to make whitespace allowed, however:
'/^\s*(...)\s*$/'
You can also use "\d" in place of "[0-9]:
'/(\d+(?:\.\d*)?)([TMB])/'
Case insensitivity:
'/.../i'
...
print("$s -> " . $matches[1] * $factors[strtoupper($matches[2])]);
Optional factor:
'/([0-9]+(?:\.[0-9]*)?)([TMB])?/'
$value = $matches[1];
$factor = isset($matches[2]) ? $factors[$matches[2]] : 1;
$output = $value * $factor;
Use printf to control output formatting:
print($value) -> 1.57E+10
printf("%f", $value); -> 15700000000.000000
printf("%.0f", $value); -> 15700000000
Stephan202 recommended a clever trick, put the "?" (optional) within the parens and you're guaranteed a match string, it just might be blank. Then you can use the blank as an array key to get the default value without having to test whether it matched or not.
Putting all that together:
$factors = Array('' => 1, 'T' => 1e3, 'M' => 1e6, 'B' => 1e9);
if (0 < preg_match('/^\s*(\d+(?:\.\d*)?)([TMB]?)\s*$/i', $s, $matches)) {
$value = $matches[1];
$factor = $factors[$matches[2]];
printf("'%s' -> %.0f", $s, $value * $factor);
}

Answer 2

Alternative version which performs a conversion on all prices found:
$multiply = array('' => 1, 'T' => 1000, 'M' => 1000000);
function convert($matches) {
global $multiply;
return $matches[1] * $multiply[$matches[3]];
}
$text = '15M 15.3M 15.3T 15';
print(preg_replace_callback('/(\d+(\.\d+)?)(\w?)/', 'convert', $text));

Answer 3

In Perl:
#!/usr/bin/perl
use strict;
use warnings;
my %factors = (
B => 1_000_000_000,
M => 1_000_000,
T => 1_000,
);
while ( <DATA> ) {
next unless m{
^
(?<number> -? [0-9]+ (?:\.[0-9]+)? )
(?<factor>B|M|T)?
$
}x;
my $number = $+{factor}
? sprintf( '%.0f', $+{number} * $factors{ $+{factor} } )
: $+{number}
;
print $number, "\n";
}
__DATA__
15M
15B
15T
15.3M
15.3B
15.3T
15
15.3
Output:
C:\Temp> z
15000000
15000000000
15000
15300000
15300000000
15300
15
15.3

Answer 4

Have no idea how to make it in one regular expression, here is my try in Ruby.
#!/usr/bin/ruby
def trans s
h={"M"=>10**6 ,"T"=>10**3}
a = s.split(/[MT]/) + [s.split("").last]
(a[0].to_f * h[a[1]]).to_i
end
puts trans "15M"
puts trans "15.3M"
puts trans "15T"
puts trans "15.3T"